What Is The Tan (a + b) Formula?
The Tan (a + b) formula is the compound-angle identity for the tangent of a sum. For any two angles $a$ and $b$ whose tangents exist, it states:
$$\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$$
Read it as a recipe. The numerator adds the two tangents. The denominator subtracts their product from $1$. The single most common wrong guess, $\tan(a+b) = \tan a + \tan b$, throws the denominator away and is false for almost every pair of angles.
The identity is one of the sum and difference identities, and it works in both degrees and radians. In radians the same statement reads $\tan\left(a+b\right)$ with $a$ and $b$ measured in radians rather than degrees. A quick numerical check with $a = 45^\circ$ and $b = 30^\circ$ gives $\tan 75^\circ = 2 + \sqrt{3} \approx 3.7321$, which we derive in full below.
How Is The Tan (a + b) Formula Derived From Sine And Cosine?
Tangent is a ratio, $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, so the sum formula follows directly from the sine and cosine sum formulas. Start with the two you already know:
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
$$\cos(a+b) = \cos a \cos b - \sin a \sin b$$
Write the tangent of the sum as their quotient:
$$\tan(a+b) = \frac{\sin(a+b)}{\cos(a+b)} = \frac{\sin a \cos b + \cos a \sin b}{\cos a \cos b - \sin a \sin b}$$
Now the key move. Divide every term, top and bottom, by $\cos a \cos b$:
$$\tan(a+b) = \frac{\dfrac{\sin a \cos b}{\cos a \cos b} + \dfrac{\cos a \sin b}{\cos a \cos b}}{\dfrac{\cos a \cos b}{\cos a \cos b} - \dfrac{\sin a \sin b}{\cos a \cos b}}$$
Each small fraction simplifies. In the numerator, $\dfrac{\sin a}{\cos a} = \tan a$ and $\dfrac{\sin b}{\cos b} = \tan b$. In the denominator, the first term becomes $1$ and the second becomes $\tan a \tan b$:
$$\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$$
That is the whole derivation. Dividing by $\cos a \cos b$ is the one step worth remembering, because it is what turns a sine-and-cosine expression into a pure tangent one.
When Is Tan (a + b) Undefined?
The formula fails at exactly one place: when the denominator hits zero. Setting $1 - \tan a \tan b = 0$ gives the undefined condition:
$$\tan a \tan b = 1$$
When that happens, $a + b$ lands on an angle where tangent itself does not exist, such as $90^\circ$ (or $\tfrac{\pi}{2}$ radians). The clearest case is $a = b = 45^\circ$: both tangents equal $1$, their product is $1$, and $a + b = 90^\circ$, where $\tan 90^\circ$ is undefined. See tan 90 degrees for why that value has no finite answer.
The honest reading matters here. A vanishing denominator is not a signal to write "infinity" as a tidy result. It is a signal that the tangent of that particular sum simply does not exist, and the sine-cosine form $\dfrac{\sin(a+b)}{\cos(a+b)}$ shows why: the cosine underneath has gone to zero.
How Do You Use The Tan (a + b) Formula? (Worked Examples)
The formula shines when an awkward angle splits into two friendly ones. Each example below is checked numerically.
Example 1: Find $\tan 75^\circ$.
Split $75^\circ$ as $45^\circ + 30^\circ$, using tan 45 degrees $= 1$ and tan 30 degrees $= \dfrac{1}{\sqrt{3}}$.
$$\tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$$
Multiply top and bottom by $\sqrt{3}$:
$$\tan 75^\circ = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$
Rationalise by multiplying by $\dfrac{\sqrt{3} + 1}{\sqrt{3} + 1}$:
$$\tan 75^\circ = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$
Final answer: $\tan 75^\circ = 2 + \sqrt{3} \approx 3.7321$.
Example 2: Find $\tan 105^\circ$.
Split $105^\circ$ as $60^\circ + 45^\circ$, with tan 60 degrees $= \sqrt{3}$:
$$\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3}\cdot 1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$$
Rationalising gives $\tan 105^\circ = -(2 + \sqrt{3}) \approx -3.7321$. The negative sign is correct: $105^\circ$ sits in the second quadrant, where tangent is negative.
Example 3: Combine two tangents directly.
Suppose $\tan a = \dfrac{1}{2}$ and $\tan b = \dfrac{1}{3}$. Then:
$$\tan(a+b) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}\cdot\frac{1}{3}} = \frac{\frac{5}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$$
So $a + b = 45^\circ$. Two messy-looking tangents collapse to a clean answer, which is exactly the kind of simplification the identity was built for.
Where Does Tan (a + b) Sit On The Unit Circle?
Tangent has two anchors, and both agree. In a right triangle, $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$. On the unit circle, the point at angle $\theta$ is $(\cos\theta, \sin\theta)$, and tangent is the slope of the line from the origin to that point, $\dfrac{\sin\theta}{\cos\theta}$. See sin cos tan for the shared definition.
Adding angles is a rotation. Turning by $a$ and then by $b$ rotates the terminal ray by $a + b$, and the tangent of that final ray is the slope you read off. The sum formula is the algebra that computes the new slope from the two original ones, without measuring the combined angle directly.
What Are The Related Tangent Values And The Companion Formula?
The building-block angles below feed almost every sum you will meet. All tangents are exact, with decimals to four places.
Table: Tangent of the common building-block angles used with the sum formula.
Angle | Radians | $\tan$ (exact) | $\tan$ (decimal) |
|---|---|---|---|
$0^\circ$ | $0$ | $0$ | $0.0000$ |
$30^\circ$ | $\tfrac{\pi}{6}$ | $\tfrac{1}{\sqrt{3}}$ | $0.5774$ |
$45^\circ$ | $\tfrac{\pi}{4}$ | $1$ | $1.0000$ |
$60^\circ$ | $\tfrac{\pi}{3}$ | $\sqrt{3}$ | $1.7321$ |
$75^\circ$ | $\tfrac{5\pi}{12}$ | $2 + \sqrt{3}$ | $3.7321$ |
$90^\circ$ | $\tfrac{\pi}{2}$ | undefined | undefined |
For a full grid of ratios, see the trigonometric table.
The difference formula is the twin of the sum, with two signs flipped:
$$\tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$$
Notice the pattern: the numerator copies the sign inside the bracket, and the denominator takes the opposite sign. That single rule keeps the two apart. As a check, $\tan 15^\circ = \tan(45^\circ - 30^\circ) = 2 - \sqrt{3} \approx 0.2679$. The companion page tan (a − b) works this side in full, and setting $a = b$ turns the sum formula into the double-angle formula for tangent.
Why Is The Denominator There At All?
The denominator is the whole difference between tangent and a quantity you could just add. Here is the reasoning, in pieces.
Tangent is a ratio, not a length. Because $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, combining two of them means combining two fractions, and fractions do not add by adding numerators alone.
The cosines interact. In $\cos(a+b) = \cos a \cos b - \sin a \sin b$, the two angles mix through the product $\sin a \sin b$. Dividing through by $\cos a \cos b$ turns that mixing term into $\tan a \tan b$, which is exactly the denominator.
It guards the undefined case. When $\tan a \tan b = 1$, the denominator is zero, and the formula refuses to return a value precisely because $\tan(a+b)$ does not exist there. A formula without the denominator would hide that.
So the denominator is not decoration. It carries the interaction between the two angles and the warning about where the answer breaks down.
Who Discovered The Angle-Sum Formulas?
Long before anyone wrote $\tan(a+b)$, astronomers needed to add angles to predict where planets would appear. The tool they built was a table of chords, and from it the addition formulas were born.
Two more figures carried the idea forward:
Aryabhata (476–550 CE, India) compiled one of the earliest sine (jya) tables, giving later mathematicians the accurate values that addition formulas rely on.
Al-Battani (around 858–929 CE, Mesopotamia) refined these tables and worked with the ratio we now call tangent, sharpening the astronomy that the compound-angle rules serve.
Where Is The Tan (a + b) Formula Used In The Real World?
The sum formula appears wherever two angles or two slopes have to be combined into one.
Surveying and construction: stacking one incline on another, as when a ramp turns partway up, gives a combined tilt whose tangent is exactly $\tan(a+b)$.
Electronics and AC signals: the phase of a combined alternating current adds two angles, and tangent relations convert between the phase and the ratio of components.
Robotics and computer graphics: composing two rotations of a joint or a camera adds their angles, and tangent identities keep the slope calculations exact rather than approximate.
Navigation: adjusting a bearing by a known turn combines two directions, and the tangent of the result follows the same rule.
One identity, built by astronomers, now quietly runs surveying instruments, power grids, animation engines, and navigation systems. The mathematics of adding angles is the same across every one of them.
What Are The Most Common Mistakes With Tan (a + b)?
These four errors account for most lost marks on compound-angle questions, and each has a clean fix.
Distributing tangent over the sum.
Where it slips in:
A student writes $\tan(a+b) = \tan a + \tan b$, treating tangent like a term that spreads across a bracket.
Don't do this:
Tangent does not distribute. $\tan(a+b)$ is almost never equal to $\tan a + \tan b$.
The correct way:
Use the full identity, numerator over denominator: $\tan(a+b) = \dfrac{\tan a + \tan b}{1 - \tan a \tan b}$.
Borrowing the wrong denominator sign.
Where it slips in:
A student recalls the sum formula while doing a difference, and writes $\tan(a-b) = \dfrac{\tan a - \tan b}{1 - \tan a \tan b}$.
Don't do this:
Do not reuse the minus in the denominator for a difference.
The correct way:
Numerator and denominator carry opposite signs. Sum: $1 - \tan a \tan b$. Difference: $1 + \tan a \tan b$.
Mixing degrees and radians on the calculator.
Where it slips in:
A student computes $\tan 45^\circ$ with the calculator set to radian mode and gets a number nowhere near $1$.
Don't do this:
Do not trust a decimal without checking the angle MODE first.
The correct way:
Set the mode to match the angle. In degree mode $\tan 45^\circ = 1$; in radian mode use $\tan\tfrac{\pi}{4} = 1$.
Calling the undefined case "infinity".
Where it slips in:
When $1 - \tan a \tan b = 0$, a student writes the answer as $\infty$ and moves on.
Don't do this:
Do not record a division by zero as a finite or infinite value.
The correct way:
State that $\tan(a+b)$ is undefined there, because $a + b$ reaches an angle like $90^\circ$ where tangent does not exist.
Practice Problems On Tan (a + b)
Work each one, then check against the answer.
Find $\tan(a+b)$ when $\tan a = 3$ and $\tan b = \tfrac{1}{2}$.
(Answer: $\dfrac{3 + \frac{1}{2}}{1 - 3\cdot\frac{1}{2}} = \dfrac{3.5}{-0.5} = -7$.)Evaluate $\tan 75^\circ$ using $45^\circ + 30^\circ$.
(Answer: $2 + \sqrt{3} \approx 3.7321$.)Show that $\tan(45^\circ + 45^\circ)$ is undefined.
(Answer: denominator $1 - (1)(1) = 0$, so it is undefined; indeed $\tan 90^\circ$ does not exist.)Simplify $\tan(x + 45^\circ)$.
(Answer: $\dfrac{\tan x + 1}{1 - \tan x}$.)Find $\tan 105^\circ$ using $60^\circ + 45^\circ$.
(Answer: $-(2 + \sqrt{3}) \approx -3.7321$.)If $\tan a = \tfrac{1}{2}$ and $\tan b = \tfrac{1}{3}$, find $a + b$.
(Answer: $\tan(a+b) = 1$, so $a + b = 45^\circ$.)
Where Should You Go Next After Tan (a + b)?
The sum formula is one door into compound-angle trigonometry, and several rooms open from here.
Sum and difference identities. The full family for sine, cosine, and tangent, all proven from the same starting point.
Trigonometric identities. The wider toolkit, from Pythagorean to double-angle, that these formulas feed into.
Tangent function. A deeper look at tangent itself, its graph, and where it is undefined.
The angle-sum identities appear in India's NCERT Class 11 (Chapter 3, Trigonometric Functions) and in the United States under the Common Core high-school standard CCSS.HSF-TF, so the same rule serves students across both systems. If your child is building these foundations, a live Bhanzu trainer teaches the compound-angle formulas starting from the "why" (rotation and the sine-cosine proof) in the Bhanzu trigonometry program.
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